At equilibrium, a reaction vessel contains 4.50 atm of Br₂ and 1.10 atm of NBr₃. According to the reaction: 2 NBr₃ (g) ⇌ N₂ (g) + 3 Br₂ (g) Kp = 4.8 Determine the equilibrium partial pressure of N₂.

Answer 1

2NBr3(g) <===> N2(g) + 3Br2(g)

Kp = (P_N2)(P_Br2)³ / P(_NBr3)²

4.8 = (P_N2)(4.50)³ / (1.10)²

5.808 = 91.125 (P_N2)

P_N2 = 0.0638 atm